Answer to Question #238326 in Mechanical Engineering for Richard

Question #238326
Question 2
A flat belt is required to transmit 22 kW from a 250 mm diameter pulley running at 1450
rpm to a 355 mm diameter pulley. The proposed belt is 3.5 mm thick.
The other system specifications are as follows:
 Coefficient of friction 0.7,
 Density of the belt is 1100 kg/m3
 Maximum permissible stress is 7 MPa.
 The distance between the shafts is 1.8 m.
2.1 Calculate the width required. [15]
1
Expert's answer
2021-09-17T02:35:58-0400

Maximum tension T,

"P = \\frac{2\\pi N T}{1000}"


"T =\\frac{22000\\times1000}{2\\pi\\times1450} = 2414.76 N"


Width of belt b,

"b = \\frac{T}{\\sigma\\times t}= \\frac{2414.76}{7\\times 3.5}=98.56" mm 






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